An engine using 1 mol of an ideal gas ini- tially at 17.8 L and 283 K performs a cycle consisting of four steps:
1) an isothermal expansion at 283 K from 17.8 L to 35.1 L;
2) cooling at constant volume to 125 K ;
3) an isothermal compression to its original volume of 17.8 L; and
4) heating at constant volume to its original temperature of 283 K .
Find its efficiency. Assume that the heat capacity is 21 J/K and the univer- sal gas constant is 0.08206 L · atm/mol/K = 8.314 J/mol/K.
To find the efficiency, we need to find the total work done in the cycle and the total heat absorbed.
The work done in the first step (the isothermal expansion) is given by
W₁ = nRT₁ * ln(V₂ / V₁)
where n is the number of moles, R is the universal gas constant, T₁ is the initial temperature, V₁ is the initial volume, and V₂ is the final volume.
In the second step (cooling at constant volume), the work done is zero since the volume doesn't change. (W₂ = 0)
The work done in the third step (isothermal compression) is given by
W₃ = nRT₃ * ln(V₁ / V₂), where T₃ is the temperature during compression.
In the fourth step (heating at constant volume), the work done is again zero since the volume doesn't change. (W₄ = 0)
The total work done in the cycle is W = W₁ + W₂ + W₃ + W₄
Now we find the heat absorbed in each step.
For the first step (isothermal expansion),
Q₁ = W₁ = nRT₁ * ln(V₂ / V₁)
For the second step (cooling at constant volume),
Q₂ = - nCvΔT₂, where Cv is the heat capacity at constant volume.
For the third step (isothermal compression),
Q₃ = W₃ = nRT₃ * ln(V₁ / V₂)
For the fourth step (heating at constant volume),
Q₄ = nCvΔT₄
The total heat absorbed in the cycle is Q = Q₁ + Q₂ + Q₃ + Q₄
Finally, the efficiency is given by
η = W / Q
Now we plug in the values to find the efficiency:
n = 1 mol
R = 8.314 J/mol/K
T₁ = 283 K
V₁ = 17.8 L
V₂ = 35.1 L
T₃ = 125 K
Cv = 21 J/K
W₁ = (1 mol)(8.314 J/mol/K)(283 K) * ln(35.1 L / 17.8 L) ≈ 5538 J
W₃ = (1 mol)(8.314 J/mol/K)(125 K) * ln(17.8 L / 35.1 L) ≈ -2522 J
W = 5538 J + 0 - 2522 J + 0 = 3016 J
Q₁ = 5538 J
Q₂ = -(1 mol)(21 J/K)(283 K - 125 K) ≈ -3324 J
Q₃ = -2522 J
Q₄ = (1 mol)(21 J/K)(283 K - 125 K) ≈ 3324 J
Q = 5538 J - 3324 J - 2522 J + 3324 J = 3016 J
η = W / Q = 3016 J / 3016 J = 1 or 100%
The efficiency of the engine is 100%.